X^2+4y^2=40

Solution for X^2+4y^2=40 equation:

X^2+4X^2=40

We move all terms to the left:

X^2+4X^2-(40)=0

We add all the numbers together, and all the variables

5X^2-40=0

a = 5; b = 0; c = -40;
Δ = b2-4ac
Δ = 02-4·5·(-40)
Δ = 800
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{2}}{2*5}=\frac{0-20\sqrt{2}}{10}
=-\frac{20\sqrt{2}}{10}
=-2\sqrt{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{2}}{2*5}=\frac{0+20\sqrt{2}}{10}
=\frac{20\sqrt{2}}{10}
=2\sqrt{2}
$